EXAM 3 MAT 167 Calculus I Spring is a composite function of two functions y = e u and u = 4 x + x 2. By the. dy dx = dy du = e u x + 2x.

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1 EXAM MAT 67 Calculus I Spring 20 Name: Section: I Each answer must include either supporting work or an explanation of your reasoning. These elements are considered to be the main part of each answer and will be MPORTANT: graded.. (0 pts) Differentiate each of the following functions. (a) y = e 4 x+x 2 Solution: y = e 4 x+x 2 Chain Rule, is a composite function of two functions y = e u and u = 4 x + x 2. By the dx = du du( dx ) 2 = e u x + 2x ( ) = e 4 x+x 2 2 x + 2x. (b) y = ln x Solution : y = ln x is a composite function of y = ln u and u = x. By the Chain Rule, dx = du du dx = u (x2 ) = x2 x = x. Solution 2: By a logarithmic identity, y = ln x = ln x. So, dx = x.

2 EXAM MAT 67 Calculus I 2 (c) y = tan (ln x) Solution: y = tan (ln x) is a composite function of y = tan u and u = ln x. By the Chain Rule, dx = du du dx = + u 2 x = x[ + (ln x) 2 ]. 2. (0 pts) Use logarithmic differentiation to find the derivative of y = x x. Solution: By a logarithmic identify, The derivative of the LHS is y Rule, dx ln y = ln x x = x ln x. by the Chain Rule. The derivative of the RHS is by the Product d dx x ln x = dx dx ln x + x d dx ln x = ln x + x x = ln x +. Thus, Hence, y dx = ln x +. dx = y(ln x + ) = xx (ln x + ).

3 EXAM MAT 67 Calculus I. (0 pts) Find the absolute maximum and the absolute minimum values of f(x) = x x + on the closed interval [0, ]. Solution: First we find the critical points of f(x) by solving the equation f (x) = 0. Since f (x) = x 2 = (x 2 ) = (x + )(x ), x =, are the critical points of f(x). Now f( ) = and f() =. On the other hand, f(0) = and f() = 9. These are all the possible candidates for the absolute maximum and the absolute minimum values. Therefore, the absolute maximum value of f(x) is f() = 9 and the absolute minimum value of f(x) is f() =. 4. (40pts) Let f(x) = x 4 8x Answer the following questions. (a) Find the intervals on which f(x) is increasing or decreasing. Solution: First we find the critical points of f(x) by solving the equation f (x) = 0: f (x) = 4x 6x = 4x(x 2 4) = 4x(x + 2)(x 2) = 0. So x = 2, 0, 2 are critical points of f(x). By using the test-point method we obtain the following table: x x < < x < < x < 2 2 x > 2 f (x) f(x) f( 2) = 0 f(0) = 6 f(2) = 0 Hence, f(x) is increasing on the intervals ( 2, 0) (2, ) and is decreasing on the intervals (, 2) (0, 2). (b) Find and identify all local maximum and local minimum values of f(x). Solution : From the table in part (a), f( 2) = f(2) = 0 is a local minimum value and f(0) = 6 is a local maximum value by the First Derivative Test. Solution 2: f (x) = 2x 2 6. Since f (0) = 6 < 0 and f ( 2) = f (2) = 2 > 0, by the Second Derivative Test, f(0) = 6 is a local maximum value and f( 2) = f(2) = 0 is a local minimum value.

4 EXAM MAT 67 Calculus I 4 (c) Find the intervals on which the graph of f(x) is concave up or concave down. Find all points of inflection. First we need to solve the equation f (x) = 0: f (x) = 2x 2 6 = 4(x 2 4) = 4( x + 2)( x 2) = 0. So f (x) = 0 at x = ± 2. By using the test-point method we find the following table: x x < < x < 2 2 x > 2 f (x) + ( 0 ) ( 0 ) + f(x) f 2 = 64 2 f 9 = 64 9 ( ) ( ) The graph of f(x) is concave on the intervals, 2 2, and is concave down on the ( ( ) ( ) interval 2 2, ). The points of inflection are 2, 64 2 and. (d) Use the information obtained from parts -, sketch the graph of y = f(x). Specify the critical points and the points of inflection on your graph. (Do not use graphing calculator.) 9, 64 9 Figure : The graph of f(x) = x 4 8x

5 EXAM MAT 67 Calculus I 5 5. (0 pts) If 200 cm 2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box Solution: Let x and y be the length (also the width) and the height of the box, respectively. The amount of material is the same as the area of the box, so we have x 2 + 4xy = 200. The volume V of the box is V = x 2 y. From the area x 2 + 4xy = 200, we obtain y = 00 x x 4. Plug this into the volume V for y, we obtain the volume as a function of x variable only: V (x) = 00x x 4. To find the largest possible volume, we need to find the critical points first: V (x) = 00 4 x2 = 0. This equation has solutions x = ±20. Since x is a length, it should be positive, so x = 20 cm and thereby y = 0 cm. Therefore, the largest possible volume would be V = 4, 000cm. Note that V (x) = x > 0, so by the Second Derivative Test, V (20) = 4, 000 is really the largest possible 2 volume. 6. (20 pts) [Optional, Bonus Questions] Find the limit using l Hôpital s rule. (a) lim t 0 e t t Solution: The limit is a 0 0 type indeterminate form, so by l Hôpital s rule, e t lim t 0 t t 0 (e t ) t t 0 e t =.

6 EXAM MAT 67 Calculus I 6 (b) lim xx Solution: The limit is 0 0 type indeterminate form. Let y = x x. Then lim ln y ln xx x ln x by a logarithmic identity. The last limit is now a 0 type indeterminate form. It can be converted to an ln x type indeterminate form by writing x ln x = : x Now we have So lim y = e0 =, i.e. ln x lim x ln x x lim xx =. (ln x) ( x) (by l Hôpital s rule) x x 2 x = lim x = 0. lim ln y = ln( lim y) = 0.